Question

Two grapes are given equal charges and held apart at a distance of 1.3 m. They experience a repulsive force of 2.2 N. Find the magnitude of the charge on each grape.

Answer 1

Explanation:

Using Coulomb's law, the relation between force and charge is as follows.

              F = $k \frac{q_{1}q_{2}}{r^{2}}$

In the given case,  $q_{1} = q_{2}$ = q

Hence,

             F = $k \frac{q^{2}}{r^{2}}$

         $q^{2} = \frac{F \times r^{2}}{k}$

Squaring on both the sides, we get

             q = $r \times \sqrt{\frac{F}{k}}$

                = $1.3 m \times \sqrt{\frac{2.2 N}{9 \times 10^{9}}}$

               = $6.422 \times 10^{-5}$ C

               = $64.22 \times 10^{-6}\mu C$

Thus, we can conclude that magnitude of the charge on each grape is $64.22 \times 10^{-6}\mu C$.